Most tested MM5.3

Solving Exponential Equations

This topic covers solving equations where the unknown variable is in the exponent, such as 3^x = 81. The key technique is using logarithms to manipulate the equation and isolate the variable.

Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.

Key points

  • To solve a basic exponential equation like a^x = b, take logarithms of both sides to 'bring down' the exponent using the power law: x log(a) = log(b).
  • Strategically choose the base of your logarithm. Taking log base 'a' is often simplest, as loga(a^x) simplifies directly to x.
  • Be alert for 'disguised quadratics'. Equations like 4^x - 6(2^x) + 8 = 0 can be transformed into a quadratic by substituting y = 2^x, giving y2 - 6y + 8 = 0.
  • After solving for the substitution variable (e.g., 'y'), you must substitute back to find the final value(s) for the original variable (e.g., 'x').
  • Always check that your solutions are valid. The argument of a logarithm must be positive, and a^x (for a > 0) is always positive, so reject any negative solutions for a substitution like y = a^x.

Formulae

logb(x^k) = k × logb(x)

This is the essential power law used to move a variable from the exponent to the main line of an equation, allowing you to solve for it.

y = a^x ≤> x = loga(y)

For converting between exponential and logarithmic forms, which is useful for solving the final step of an equation like 2^x = 5.

Definitions

Logarithm
The power to which a base number must be raised to get another number. The statement y = logb(x) is equivalent to b^y = x.

Worked example

Find the real solution(s) to the equation 4^x - 3 × 2^(x+1) + 8 = 0.

  1. 1

    First, simplify the terms to have a common exponential base.

    Note that 4^x = (22)^x = 2^(2x) and 2^(x+1) = 2^x × 21 = 2 × 2^x.

  2. 2

    Substitute these into the equation:

    (2^x)2 - 3 × (2 × 2^x) + 8 = 0.

  3. 3

    Simplify the middle term:

    (2^x)2 - 6 × (2^x) + 8 = 0.

  4. 4

    This is a quadratic in the form of 2^x.

    Let y = 2^x

    The equation becomes:

    y2 - 6y + 8 = 0.

  5. 5

    Factorise the quadratic:

    (y - 4)(y - 2) = 0.

    This gives two possible solutions for y:

    y = 4 or y = 2
  6. 6

    Solve for x by substituting back.

    If y = 4, then 2^x = 4, which means x = 2
    If y = 2, then 2^x = 2, which means x = 1
  7. 7

    Both solutions are valid.

Answer: x = 1, 2

Common mistakes

  • ×Errors in applying index laws before you even start with logarithms. For example, incorrectly handling 2^(x+1) and writing it as 2^x + 2.
  • ×Making arithmetic mistakes when factorising or using the quadratic formula, especially with negative signs.
  • ×Stopping after solving for the substituted variable (e.g., finding y = 4 and y = 2) and forgetting to convert back to find the values for x.
  • ×Incorrectly applying log rules, for instance, attempting to split log(A+B) into log(A) + log(B), which is not a valid identity.

No-calculator tips

  • ✓Before using logs, always check if the numbers are simple integer powers of each other. For 2^x = 32, recognise that 32 = 25, so x=5 immediately.
  • ✓Memorise key log identities to solve final steps quickly: logb(1) = 0 and logb(b) = 1. If you get to 5^x = 1, you know x=0 without any further calculation.
  • ✓When solving a 'hidden quadratic', double-check your factorisation. The two numbers must multiply to give the constant term and add to give the coefficient of the middle term. This simple check catches most arithmetic slips.

Read this topic in the official UAT-UK ESAT guide →

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